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3 years ago

A car decelerates at a rate of 3.8 m/s2 for 12 seconds. The car’s initial speed was 65 m/s. What was the car’s final speed?

1 answer:

4 0

For an uniformly accelerated motion, the final velocity is given by:
$v_f = v_i +at$
where vi is the initial speed, a is the acceleration, and t the time interval.
In our problem, $v_i=65 m/s$ , $a=-3.8 m/s^2$ (with negative sign, because the car is decelerating), and $t=12 s$ . Substituting,
$v_f = 65 m/s+(-3.8 m/s^2)(12 s)=19.4 m/s$

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

3 years ago

A 1,492.3-kg airplane travels down the runway. Each of its four engines provides a force of

Fittoniya [83]

The acceleration of the air plane is $3.879 \mathrm{m} / \mathrm{s}^{2}$

<u>Explanation:</u>

Given:

The mass of the air plane = 1492.3 kg

Force of each four engine = 1447.5 N

So, the total force of four engines can be calculated as = 4(1447.5) = 5790 N

The force that acts on the object is equal to the product of mass (m) and its acceleration. It can express by the below formula,

$\text {Force }(F)=m \times \text { acceleration }(a)$

The above equation can be written as below to find acceleration,

$a=\frac{F}{m}$

Now. Substitute the given values, we get,

$a=\frac{5790}{1492.3}=3.879 \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago

If ice has a density of.92g/cm, what is the volume of 1.8 kilograms of ice?

steposvetlana [31]

Given:
Density = .92 g / cm³
Volume = 1.8 kg
= (1.8 * 1000) grams
= 1800 grams
Now,
$Density (d) = \frac{Mass (m)}{volume(v)}$

$volume (v)= \frac{mass(m)}{density(d)}$

$volume (v)= \frac{1800}{.92}$

$volume (v)= 1956.521739$

$volume (v)= 1956.5~ cm^{3}$

So, the volume of 1.8 kg of ice is 1956.5 cm³

5 0

3 years ago

In thermodynamics, work is typically done by

kondor19780726 [428]

The answer is gases

3 0

3 years ago

An eagle carries a fish up 50 m into the sky using 90 N of force. How much work did the eagle do on the fish? (Work: W = Fd)

Leno4ka [110]

<span>Work: W = Fd. 50(distance) multiplied by 90(force) would equal 4500 J or, answer D</span>

4 0

3 years ago

Read 2 more answers